# Difficult coupled DE system.

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#### charlottewill

##### New member
We've been asked to solve this system of equations

dW/dt = = αW−βV
dV dt = −γV+δW

by proposing that

W (t) = A sin(ωt) + B cos(ωt)
V (t) = C sin(ωt) + D cos(ωt).

so I've differentiated both W(t) and V(t) to give W'(t) = ωA cos(ωt) - ωB sin(wt) and V'(t) = ωC cos(ωt) - ωD sin(wt) and replaced with the dW/dt and dV/dt in the system to give the expressions

ωA cos(ωt) - ωB sin(wt) = αW−βV

ωC cos(ωt) - ωD sin(wt) = −γV+δW

and now I've replaced the W and V's on the RHS with W (t) = A sin(ωt) + B cos(ωt)
V (t) = C sin(ωt) + D cos(ωt) (originally given in the question) and I've factored out the constant and the parameters of cos(ωt) and sin (ωt) to give two simultaneous equations equal to zero which are

(ωA-αB+βD) cos(ωt) + (βC-αA-ωB) sin(ωt) = 0

(ωC+γD-δB) cos(ωt) + (γC-ωD-δA) sin(ωt) = 0

So how do I find the value of the constants A,B,C and D in terms of the parameters α,β,γ,δ ω?

Thanks,

Charlotte

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#### Opalg

##### MHB Oldtimer
Staff member

We've been asked to solve this system of equations

dW/dt = = αW−βV
dV dt = −γV+δW

by proposing that

W (t) = A sin(ωt) + B cos(ωt)
V (t) = C sin(ωt) + D cos(ωt).

so I've differentiated both W(t) and V(t) to give W'(t) = ωA cos(ωt) - ωB sin(wt) and V'(t) = ωC cos(ωt) - ωD sin(wt) and replaced with the dW/dt and dV/dt in the system to give the expressions

ωA cos(ωt) - ωB sin(wt) = αW−βV

ωC cos(ωt) - ωD sin(wt) = −γV+δW

and now I've replaced the W and V's on the RHS with W (t) = A sin(ωt) + B cos(ωt)
V (t) = C sin(ωt) + D cos(ωt) (originally given in the question) and I've factored out the constant and the parameters of cos(ωt) and sin (ωt) to give two simultaneous equations equal to zero which are

(ωA-αB+βD) cos(ωt) + (βC-αA-ωB) sin(ωt) = 0

(ωC+γD-δB) cos(ωt) + (γC-ωD-δA) sin(ωt) = 0

So how do I find the value of the constants A,B,C and D in terms of the parameters α,β,γ,δ ω?
That last line should read: So how do I find the value of the constants A,B,C, D and ω in terms of the parameters α,β,γ,δ? In other words, ω is not one of the fixed parameters, it is up for grabs along with A,B,C and D.

Next, you want those two equations to hold for all values of t. The only way for that to happen is if all those coefficients of cos(ωt) and sin(ωt) are zero. Thus you are looking at the system of equations \begin{aligned}\omega A - \alpha B \phantom{{}+\gamma C} +\beta D &=0 \\ -\alpha A -\omega B +\beta C \phantom{{}+\gamma C} &=0 \\ -\delta B + \omega C + \gamma D &=0 \\ -\delta A \phantom{{}+\gamma B} +\gamma C -\omega D &=0. \end{aligned}

The is a set of four homogeneous linear equations in A,B,C and D, so normally the only solution will be when they are all 0. But for some values of the (variable) parameter $\omega$ there will be a nontrivial solution.

The details of solving that set of equations look a bit gruesome, but it looks as though that is what you are expected to do.

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